Question

A ball of mass $0.9\text{kg}$ is projected with an initial velocity of $10\text{m/s}$ at an angle of ${37}^{\circ }$ with the horizontal. After $0.6\text{sec}$, the gravitational field vanishes and a force of constant magnitude is applied on the ball afterwards, the force being always perpendicular to the direction of motion till it strikes the ground. When it strikes the ground, it is moving vertically downwards. Choose the correct option. (Take $g=10{\text{m/s}}^2$)

• A. Path of projectile after $0.6\text{sec}$ is parabolic
• B. Path of ball after $0.6\text{sec}$ will be circular and radius of circle will be $1.8\text{m}$
• C. Magnitude of the constant force applied is $32\text{N}$
• D. Speed during circular motion will be $8\text{m/s}$

Answer 1

Answer: (B), (C), (D)

Speed during circular motion will be $8\text{m/s}$


Time of flight $T=\frac{2u\sin \theta }{g}$
$=\frac{2\times 10\times \frac{3}{5}}{g}=1.2\text{s}$
$\therefore t_1=\frac{T}{2}=0.6\text{s}$ will be the time of ascent of the ball, hence at this time it will reach maximum height $H$.
$H=\frac{u^2{\sin }^2\theta }{2g}=\frac{100\times {\left(\frac{3}{5}\right)}^2}{2\times 10}=\frac{36}{20}=1.8\text{m}$

Let force $\left(F\right)$ of constant magnitude act perpendicular to the velocity of the ball always after reaching maximum height and gravity is absent throughout.
$\Rightarrow$ Ball will perform uniform circular motion about point $\text{O}$, with constant speed $u_x=10\cos {37}^{\circ }=8\text{m/s}$


Since $F$ provides centripetal force,
$F=\frac{m{u}_x^2}{R}$ where $R=H$
$F=\frac{m{v}^2}{H}$
$\therefore F=\frac{0.9\times 8^2}{1.8}=32\text{N}$
Hence, option (b), (c) and (d) are correct.